Question

If $\tan \theta +\sin \theta =m$ and $\tan \theta -\sin \theta =n$, then prove that $m^2-n^2=4\sqrt{mn}$.

• A. $4\sqrt{\frac{m}{n}}$
• B. $4\sqrt{\frac{n}{m}}$
• C. $4\sqrt{mn}$
• D. $4mn$

Answer 1

The correct option is C $4\sqrt{mn}$

Tanθ+sinθ=m and tanθ-sinθ=n

∴, m²-n²

=(m+n)(m-n)

=(tanθ+sinθ+tanθ-sinθ)(tanθ+sinθ-tanθ+sinθ)

=(2tanθ)(2sinθ)

=4tanθsinθ

4√mn

=4√(tanθ+sinθ)(tanθ-sinθ)

=4√(tan²θ-sin²θ)

=4√{(sin²θ/cos²θ)-sin²θ}

=4√sin²θ{(1/cos²θ)-1}

=4sinθ√{(1-cos²θ)/cos²θ}

=4sinθ√(sin²θ/cos²θ)

=4sinθ√tan²θ

=4sinθtanθ

∴, LHS=RHS (Proved)